Yet Another Power Tower

Calculus Level 4

$\huge a^{a^{a^{\cdot^{\cdot^\cdot}}}}=\frac{1}{3}$

How many positive real solutions $a$ does the above equation have?

Hint : Argue "from first principles", without using any properties of the infinite power tower you might know.

Clarification : The value of the infinite power tower $a^{a^{a^{.^{.^.}}}}$ is defined as the limit of the sequence $x_1=a, x_{n+1}=a^{x_n},$ if that limit exists.

3 Infinitely many 2 1 0

2 solutions

Otto Bretscher
Feb 23, 2016

My solution will summarize the discussion we had with @Brian Charlesworth , @Andreas Wendler et alii. See Andreas' solution.

If we had a solution $a$ , then the equation $a^{1/3}=\frac{1}{3}$ would hold, so $a=(\frac{1}{3})^3=\frac{1}{27}$ . This shows that the given equation has at most one solution. The value $a=\frac{1}{27}$ may or may not be a solution, and there cannot be any other solutions. We need to examine whether the sequence $x_n$ associated with $a=\frac{1}{27}$ actually converges to $c=\frac{1}{3}$ .

If we define the iteration function $f(x)=a^x=(\frac{1}{27})^x$ , then we observe that $f'(x)=\ln(a)a^x$ so $f'(c)=\ln(a)a^c=\ln(a)c=-\ln(3)<-1$ . This implies, by the Mean Value Theorem, that $x_n$ fails to converge to $c$ . Indeed, if $x_n$ is sufficiently close to $c$ (but not equal to $c$ ), then $|x_{n+1}-c|$ $=|f(x_n)-f(c)|$ $=|f'(p)||x_n-c|>|x_n-c|$ , for some point $p$ between $c$ and $x_n$ : The equilibrium point $c$ actually "repels" $x_n$ .

If you draw a cobweb starting sufficiently close to $c$ , it will "spiral away" from the point $(c,c)$ , at least initially. (I wish somebody could produce a graph illustrating this behaviour; I'm not good at this.)

Thus the given equation has no solutions.

PS: This discussion applies to any $c<\frac{1}{e}$ .

Summarized this means a limit for the power tower exists only for the fixpoint a (=value of tower) if and only if $|f'(a)|<1$ with $f(x)=a^{\frac{x}{a}}$ . (You proofed clearly what I assumed in an earlier posting.)

This only happens in the valid interval [ 1/e, e ] for a. In all other cases the tower converges against at least 2 limits or perhaps not a bit.

Andreas Wendler - 5 years, 3 months ago

The condition turns out to be "if and only if $|f'(a)|\leq 1$ ". The two "Grenzfälle" need to be examined separately... we have done the case $a=e$ earlier. Also, my work above does not prove that the sequence converges when $|f'(a)|<1$ ; one needs to check that the seed $x_1=a^{1/a}$ is in the "basin of attraction". The condition $|f'(a)|<1$ guarantees convergence only locally .

"This only happens in the valid interval [ 1/e, e ] for a. In all other cases the tower converges against at least 2 limits or perhaps not a bit." There is a third possibility: If $a>e$ then the power tower of $a^{1/a}$ converges to a value other than $a$ , as in this case ; it converges to the value $b<e$ such that $a^b=b^a$ . If $0<a<\frac{1}{e}$ , then the power tower of $a^{1/a}$ approaches a two-cycle, as our collaborative solution here shows.

Otto Bretscher - 5 years, 3 months ago

If you could explain what the (c,c) value refers to, I could try graphing the trajectory

Agnishom Chattopadhyay - 5 years, 3 months ago

Thanks for the offer! The figure would involve the graph of $y=(\frac{1}{27})^x$ and the line $y=x$ , for $0\leq x \leq 1$ ; they intersect at the equilibrium point $(c,c)=(1/3,1/3)$ . Then it would be fantastic if you could draw a cobweb starting close to 1/3, maybe at 0.3 or 0.31 or 0.32... you would have to experiment a bit and see what looks best. Hopefully, we would see a limiting two-cycle emerge between circa $x=0.08$ and $x=0.77$ , with the cobweb approaching the corresponding rectangle.

If we start a cobweb at our initial point 1/27, it should approach the limiting cycle from the outside, that is, $x_n$ will be more than 0.77 or less than 0.08, in an alternating fashion.

Clearly, many people are having a hard time with these nested problems. (As of right now, this problem has 277 attempts and 56 solvers.) Some people just find the equilibrium point in the hope that it will be the solution. I trust that a good figure will go a long way in explaining things.

Thanks again!

Otto Bretscher - 5 years, 3 months ago

Can you explain what the term conweb refers to?

I think you mean something like this :

But I do not know what this is either

Agnishom Chattopadhyay - 5 years, 3 months ago

@Agnishom Chattopadhyay – Yes, that's a cobweb indeed! In a technical sense, you do this: Starting with any point $a$ on the $x$ -axis in the domain of a function $f(x)$ , draw a vertical line segment to the graph, then a horizontal line segment to the line $y=x$ , then a vertical to the graph again etc. This is a way to visualize the sequence $x_0=a, x_{n+1}=f(x_n)$ .

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher –

Larger Version

Interactive Version , with variable starting points.

Does this do the job?

Agnishom Chattopadhyay - 5 years, 3 months ago

@Agnishom Chattopadhyay – Wow, this is fantastic! Thank you so very much!

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – No problem. I had fun doing this

Agnishom Chattopadhyay - 5 years, 3 months ago

@Agnishom Chattopadhyay – I don't know how much time this takes... if it's quick, could you perhaps do one of those here ? It would be tremendously useful!

Otto Bretscher - 5 years, 3 months ago

@Otto Bretscher – I posted a comment in the solution to the problem

Agnishom Chattopadhyay - 5 years, 3 months ago

Andreas Wendler
Feb 21, 2016

1/3 is outside the domain [ 1/e, e ] valid for power towers!

True; you are invoking a powerful result due to Euler. But can you prove "from first principles" that $\frac{1}{3}$ fails to be in the range, without using Euler's result?

A lot of people naïvely "solve" these nested problems without worrying about convergence, just finding a fixed point of the iteration (as of right now, this problem was 64 attempts and 17 solvers). They might come to the mistaken conclusion that $a=\left(\frac{1}{3}\right)^3$ is the solution. Convince them that this isn't the answer!

Otto Bretscher - 5 years, 3 months ago

Why should it be in the domain [1/e,e]?

A Former Brilliant Member - 5 years, 3 months ago

In the moment I can not fix the problem exactly. But I strongly guess it has something to do with the first deviation at the fix point of the function $y=a^{\frac{1}{a}x}$ . It has value -1 for a=1/e and is less than -1 for a=1/3 but greater than -1 for possible values a<=e where deviation reaches 1.

Andreas Wendler - 5 years, 3 months ago

With the help of a little VB-application one can show that there are two limits for the tower, namely 0.078 and 0.77.

Andreas Wendler - 5 years, 3 months ago

Indeed, the problem is analogous to this , with $a=\frac{1}{27}$ instead of $a=0.05$ .

Otto Bretscher - 5 years, 3 months ago

Yet Another Power Tower

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